JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2022
- Hard
- Numerical
In a hydrogen spectrum, \(\lambda\) be the wavelength of first transition line of Lyman series. The wavelength difference will be "a \(\lambda^{\prime}\) between the wavelength of \(3^{\text {rd }}\) transition line of Paschen series and that of \(2^{\text {nd }}\) transition line of Balmer Series where \(a=........\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
For first line of Lyman
\(\frac{1}{\lambda}= R \left(1-\frac{1}{4}\right)= R \left(\frac{3}{4}\right)\)
\(\quad \lambda =\frac{4}{3 R } \ldots(1)\)
\(3^{\text {rd }} \text { line(Paschen) }\)
\(\frac{1}{\lambda_{3}}= R \left(\frac{1}{3^{2}}-\frac{1}{6^{2}}\right)=\frac{ R }{9} \times \frac{3}{4}\)
\(2^{nd}\) line(Balmer)
\(\frac{1}{\lambda_{2}}= R \left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{ R }{4} \times \frac{3}{4}\)
Thus a \(\lambda=\lambda_{3}-\lambda_{2}=\frac{12}{R}-\frac{16}{3 R}=\frac{20}{3 R}\) putting
\(a\left(\frac{4}{3 R}\right)=\frac{20}{3 R} \Rightarrow a=5\)
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